✁ 3.9
✄☎
✂
✆✝✄ ✞x(mN + n) = x(n)
(1)IDFT ✂
DFT
✟✠] ✡
IDFT ✂
[
x ( n) =
1
N
N −1
∑ X ( k )e
j
2πnk
N
,n = 0,1,...N − 1
k =0
☛☞✌✍✎ n = mN + n ✏☞✑✏
x(mN + n) =
=
N −1
1
N
∑ X (k ) exp{ j
1
N
∑ X (k ) exp( j
k =0
N −1
k =0
2π (mN + n)k
}
N
2πnk
) exp( j 2πmk )
N
✒✒✓✎ exp( j 2πmk ) = cos 2πmk + sin 2πmk = 1 ✔✂✓
1 N −1
2πnk
X (k ) exp( j
)
∑
N k =0
N
= x(n)
x(mN + n) =
✏✔✕✎IDFT ✂✆✝✄
x(mN + n) = x(n)
✖✗✘✙✚✛
✞
(4) ✜✢✣✤✥ x ( n + m) ↔ e
−j
2πk
m0
N
X (k )
✟✠] ✆✝ ✆✝✄✦✧★ ✙✚
✏ ✛ x(n + m ) ✂ DFT ✩✎
x(n) ✖
N✂
[
0
N −1
∑ x(n + m
0
)e
−j
2πnk
N
n =0
= x(m0 ) + x(1 + m0 )e
+ x ( N )e
−j
−j
2πk
N
2π ( N − m0 ) k
N
+ .... + x( N − 1)e
+ x( N + 1)e
−j
−j
2π ( N −1− m0 ) k
N
2π ( N +1− m0 ) k
N
+ ... + x( N − 1 + m0 )e
✪✫✫✬✭ x( N ) = x(0), x( N + 1) = x(1), x( N − 1 + m ) = x(m
0
N −1
= ∑ x ( n) e
−j
2π ( n − m0 ) k
N
n =0
2πm0 k N −1
j
N
∑ x ( n )e
=e
−j
2πnk
N
n =0
=e
−j
2πk
m0
N
X (k )
✱✲✳✭ x(n + m) ↔ e
−j
2πk
m0
N
X (k )
✴✵✶✷✸✹
0
− 1)
−j
2π ( N −1) k
N
✮✯✬✭✰

PART-A EXPERIMENTS USING MATLAB

Document 7473831